Answer:
b.) 4h
Explanation:
Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed of 121 km/h. At t = 0, car B is 41 km behind car A.
(a) How much farther will car A travel before car B overtakes it?km
(b) How much ahead of A will B be 30 s after it overtakes A?km
Car A will travel 150.1 km further before Car B overtakes it, and 30 seconds after Car B overtakes Car A, Car B will be 1.005 km ahead of Car A.
Explanation:This is a physics problem involving relative speed and time. We are trying to find out how much farther Car A will travel before Car B catches up, as well as Car B's lead distance 30 seconds after it overtakes Car A.
a) How much farther will car A travel before car B overtakes it?
First, we calculate the Relative Speed of the two cars: Relative Speed = Speed of B - Speed of A = 121 km/h - 95 km/h = 26 km/h. Then we find out how long it takes for Car B to close that 41 km gap at this relative speed: Time = Distance/Speed = 41km / 26km/h = 1.58 hours. Therefore, in this duration, Car A will travel Distance = Speed * Time = 95km/h * 1.58h = 150.1 km.
b) How much ahead of A will B be 30 s after it overtakes A?
Here we just calculate the distance Car B travels in 30 seconds (converted to hours). Distance = Speed * Time = 121km/h * (30s / 3600s/h) = 1.005 km.
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Car A will travel approximately 150.1 km before Car B overtakes it. Car B will be approximately 0.22 km ahead of Car A 30 seconds after overtaking. The solution involves calculating the time taken for Car B to close the initial gap and then the additional distance traveled in the subsequent 30 seconds.
Part (a)
To determine how much farther Car A will travel before Car B overtakes it, we can set up an equation based on the relative speeds of the two cars and the distance between them.
Let the time taken for Car B to overtake Car A be t hours.
In time t, Car A travels 95t km.
In time t, Car B travels 121t km.
At the point of overtaking, the distance covered by Car A plus the initial 41 km (since Car B started 41 km behind) will be equal to the distance covered by Car B:
95t + 41 = 121t
Solving for t:
121t - 95t = 41
26t = 41
t = 41/26 ≈ 1.58 hours
The distance Car A travels in this time is:
95 km/h × 1.58 hours ≈ 150.1 km
Part (b)
To find out how far ahead Car B will be 30 seconds after overtaking Car A:
Convert 30 seconds to hours:
30 seconds = 30/3600 hours = 1/120 hours.
In 1/120 hours :
Car A travels: 95 km/h × 1/120 hours ≈ 0.79 km.
Car B travels: 121 km/h × 1/120 hours ≈ 1.01 km.
The difference in distance traveled by Car B and Car A:
1.01 km - 0.79 km ≈ 0.22 km.
So, Car B will be approximately 0.22 km ahead of Car A 30 seconds after overtaking it.
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100 rad/s2. After making 2844 revolutions, its angular speed is 140 rad/s.
Complete answer
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100rad/s2. After making 2844 revolutions, its angular speed is 140rad/s
(a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?
Answer:
a) 126.59 radians per second
b) 134.1 seconds
Explanation:
We can use the rotational kinematic equations for constant angular acceleration.
a) For a) let’s use:
[tex]\omega^{2}=\omega_{0}^{2}+2\alpha\varDelta\theta [/tex] (1)
with [tex] \omega_{0}[/tex] the initial angular velocity, [tex] \omega[/tex] the final angular velocity, [tex] \alpha[/tex] the angular acceleration and [tex] \Delta \theta [/tex]the revolutions on radians (2844 revolutions = 17869.38 radians). Solving (1) for initial velocity:
[tex]\sqrt{\omega^{2}-2\alpha\varDelta\theta}=\omega_{0} [/tex]
[tex]\omega_{0}^2=\sqrt{(140)^2 -(2)(0.100)(17869.38)=126.59 \frac{rad}{s}}[/tex]
b) Knowing those values, we can use now the kinematic equation
[tex] \omega=\omega_{0}+\alpha t[/tex]
with t the time, solving for t:
[tex]t=\frac{\omega-\omega_0}{\alpha}=\frac{140-126.59}{0.1} [/tex]
[tex] t=134.1 s[/tex]
Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first and last car is moved from rest by the locomotive. Discuss the advisability of this loose coupling and slack between cars from the point of view of impulse and momentum.
Answer:
Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).
Explanation:
Why are biogeochemical cycles important? Earth is a closed system. That means the amount of rocks, metals, carbon, nitrogen, oxygen, and water on Earth.... So, it is essential that biogeochemical cycles..... these materials as they move through Earth's subsystems.
Answer:
The earth is considered to be a closed system, where the concentration of rocks, metals, C, N, O and H₂O on earth remains constant.
So, it is essential that the biogeochemical cycle renews these materials while moving through the earth subsystems.
Explanation:
The biogeochemical cycle is usually defined as the circulation of chemical components such as Carbon, Nitrogen, Phosphorous, Oxygen, rocks, and metals within the different spheres and interaction of biotic and non-biotic factors.
These components are continuously in motion from one sphere to another. For example, the organism takes up the elements through food and some of these components such as carbon is released into the atmosphere in the form of CO₂. After the death of the organisms, the decomposers feed on these elements, where some of it is consumed by them and the remaining is eliminated into the atmosphere.
So, this is how the elements are renewed, and they are conserved. The total concentration of these elements remains constant considering the earth to be a closed system.
Thus, the correct answers are given above.
Answer:
remains constant and renews
Explanation:
A child in danger of drowning in a river is being carried down-stream by a current that flows uniformly with a speed of 2.20 m/s . The child is 500 m from the shore and 1100 m upstream of the boat dock from which the rescue team sets out.
If their boat speed is 7.30 m/s with respect to the water, at what angle from the shore must the boat travel in order to reach the child?
Final answer:
To reach a child being carried by a river current, the rescue boat, with a speed of 7.30 m/s versus water, must aim at a specific angle upstream accounting for the current's speed of 2.20 m/s. The calculation involves vector addition and trigonometry to counteract the current and ensure a direct path to the child.
Explanation:
The question involves calculating the angle at which a rescue boat must travel to reach a child being carried away by a river current. The river flows at a speed of 2.20 m/s, while the boat's speed in still water is 7.30 m/s. Given the distances from the shore to the child (500 m) and from the boat dock upstream to the child (1100 m), we need to determine the angle relative to the shore for the boat to make a direct rescue.
To solve this, we need to use the concept of vector addition, where the boat's velocity vector and the river's current vector combine to create a resultant vector that points directly to the child's location. The angle can be calculated using trigonometry, particularly the tangent function. However, since the specific calculations and steps to obtain the angle are not provided in the question or the previous examples, it is crucial to understand the principles of vectors and relative motion to approach this problem.
Essentially, the rescue team needs to aim the boat at an angle upstream to counteract the river's flow. The effect of the river's current will adjust the boat's path towards the child. It's a classic application of relative motion in two dimensions, requiring careful consideration of both the magnitude and direction of the vectors involved.
A swimmer is determined to cross a river that flows due south with a strong current. Initially, the swimmer is on the west bank desiring to reach a camp directly across the river on the opposite bank. In which direction should the swimmer head?
Answer:
Swim in a strong north easterly direction so the the river carries him right across
Explanation:
The swimmer is on the west river bank with intentions of reaching the east river bank directly across him. If he just swims towards the east, the current of the river will carry him downstream landing him in a south easterly position, below where he intended to go. So the right thing to do would be to swim in a strong north easterly direction so the the river carries him right across.
To reach directly across the river from the west bank, the swimmer should head northeast to counteract the southward current of the river. This balancing act of velocities ensures they reach their desired location.
Explanation:This question deals with the concept of Relative Velocity in Physics. When a swimmer attempts to cross a river, they must take into account the current of the river pushing them downstream. The velocity of the swimmer relative to the water needs to be added to the velocity of the river to get the resultant velocity, which describes the direction the swimmer actually travels relative to the shore.
In this scenario, if the current of the river is flowing due south, the swimmer on the west bank should not aim directly for the point across the river. They should aim slightly upstream to the north to counteract the force of the current. Hence, the swimmer should head in a northeasterly direction, which when combined with the river current, will get them directly across to the eastern shore.
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If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was, how many rotations per second will result?
Answer:
There are finally 4 rotations per second.
Explanation:
If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was. We need to find the final angular velocity. It is a case of conservation of angular momentum such that :
[tex]I_1\omega_1=I_2\omega_2[/tex]
Let [tex]I_1=I[/tex] , [tex]I_2=\dfrac{I}{4}[/tex] and [tex]\omega_1=1[/tex]
So,
[tex]\omega_2=\dfrac{I_1\omega_1}{I_2}[/tex]
[tex]\omega_2=\dfrac{I\times 1}{(I/4)}[/tex]
[tex]\omega_2=4\ rev/sec[/tex]
So, there are finally 4 rotations per second. Hence, this is the required solution.
Final answer:
When a trapeze artist reduces their rotational inertia to one fourth while rotating, their rotational speed increases to four times its original rate to conserve angular momentum, resulting in 4 rotations per second.
Explanation:
The question pertains to the conservation of angular momentum, which is a principle in physics stating that if no external torque acts on an object, the total angular momentum will remain constant. For a trapeze artist or any rotating body, if the moment of inertia decreases, the rotational speed (number of rotations per second) must increase proportionally to conserve angular momentum.
If a trapeze artist is rotating once each second and contracts to reduce the rotational inertia to one fourth, her rotational speed must increase to four times what it was to conserve angular momentum. Therefore, the new rotation rate will be 4 rotations per second.
The outside of a picture frame has a length which is 3 cm more than width. The area enclosed by the outside of the picture frame is 154 square cm. Find the width of the outside of the picture frame.
Answer:
To me it would only make sense for it to be 51.33
Explanation:
w≈51.33
Length 3
Area 154
The width of the outside of the picture frame is 11 cm.
What is width?The distinction between a figure's length and width is that length denotes a figure's longer side, while width denotes its shorter side.
Given
L = w + 3
L w = 154
(w+3)w = 154
w² + 3 w - 154 = 0
(w+14)(w-11) = 0
w = 11 cm
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Which answer most completely describes the difference/s between the gravitational force and the electrostatic force? Question 6 options:
The gravitational force is much weaker than the electrostatic force
The gravitational force is weaker and can only attract objects but the electrostatic force is strong and can both attract and repel
The gravitational force occurs throughout the universe but the electrostatic force only occurs on Earth
The gravitational force can only attract but the electrostatic force can attract or repel
I believe the answer would be D
The gravitational force is weaker and can only attract objects but the electrostatic force is strong and can both attract and repel. - This is the most completely describes the differences between the gravitational force and the electrostatic force.
What are the differences between gravitational force and electrostatic force?The main differences are:
Electrostatic force acts on particles with charge, whereas gravitational force acts on particles with mass.While the electrostatic force can be either attracting or repulsive, the gravitational force is always attractive.While the electric potential can be either negative or positive, the gravitational potential is always negative.Gravitational force is weak but has low range whereas electrostatic force is strong bur short ranged.Learn more about electrostatic force here:
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A car is traveling at 22 m/s and slows to 4.9 m/s in 4.0 seconds. What is the acceleration? Give the answer to one decimal place.
Answer:
a= - 4.2 m/s².
Explanation:
Given that
u = 22 m/s
v= 4.9 m/s
t= 4 s
The average acceleration = a
We know v = u +at
v=final velocity
u=initial velocity
Now by putting the values in the above equation
4.9= 22 + a x 4
[tex]a=\dfrac{4.9-22}{4}\ m/s^2[/tex]
a= - 4.2 m/s².
Therefore the acceleration will be - 4.2 m/s².
a= - 4.2 m/s².
Negative indicates that velocity and acceleration is is opposite direction.
An elevator packed with passengers has a mass of 1950 kg.
(a) The elevator accelerates upward (in the positive direction) from rest at a rate of 2 m/s2 for 2.05 s. Calculate the tension in the cable supporting the elevator in newtons.
(b) The elevator continues upward at constant velocity for 8 s. What is the tension in the cable, in Newtons, during this time?
(c) The elevator experiences a negative acceleration at a rate of 0.45 m/s2 for 2.6 s. What is the tension in the cable, in Newtons, during this period of negative accleration?
(d) How far, in meters, has the elevator moved above its original starting point?
Answer:
a) Fa= Tension in the cable= 23010N
b) 19110N
c) 18232.5N
d=41.28m ,final velocity=0
Explanation:
a) Newtons 2nd law is given by Fnet=EF=ma
Fa= m(a + g)
Fa= 1950 (2+9.8)
Fa= 1950×11.8= 23010N
b) Fnet=0
Therefore Fb= W= 1950×9.8= 19110N
c) Fc= m(g - a)
Fc= 1950(9.8 - 0.45)
Fc= 18232.5N
d) First distance
ya= vt + 0.5at^2 = 0 + (0.5)(2)(2.05)^2
ya= 4.203m
yb= vt= 4.203×8=33.62m
yc = vt - (0.5at^2)
yc= 4.203×2.6) - (0.5×0.45×8^2)
yc = 10.93-14.4
yc =-3.46m
Dtotal = -3.46+33.62+4.203
Dtotal=41.28m
Explanation:
Below is an attachment of the solution.
One of the moons of a planet is at an average distance of 0.7 AU from the planet and its period of revolution is 27.8 earth days. What is the period of revolution of another moon of the same planet that is at 1.2 AU from in?
a. 62.3 earth days
b. 128.4 earth days
c. 224 earth days
d. 27.8 earth days
Answer:
option A
Explanation:
given,
distance of the moon 1 from planet = 0.7 AU
Time period of moon 1 = 27.8 earth days
distance of the moon 2 from the planet = 1.2 AU
time period of the moon 2 = ?
using Kepler's formula
[tex]T^2 = \dfrac{4\pi^2}{GM}\ R^3[/tex]
now,
only variable in the formula is time period and distance.
so,
[tex]\dfrac{T_2^2}{T_1^2}=\dfrac{R_2^3}{R_1^3}[/tex]
[tex]\dfrac{T_2^2}{27.8^2}=\dfrac{1.2^3}{0.7^3}[/tex]
T₂² = 3893.49
T₂ = 62.3 earth days
Hence, the correct answer is option A
You find a bag labeled "10 kg of lead-210". Its contents now weigh 1.25 kg. How many half-life periods have there been since the bag was weighed and labeled? Lead-210 is a radioisotope with a half-life of about 22 years.
Answer : The number of half-life periods will be, 3
Explanation : Given,
Initial amount of lead = 10 kg
Amount of lead after decay = 1.25 kg
Half-life = 22 years
Formula used :
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives
[tex]a_o[/tex] = Initial amount of the reactant
n = number of half lives
[tex]t_{1/2}[/tex] = half-life
Now put all the given values in the above formula, we get:
[tex]1.25=\frac{10}{2^n}[/tex]
[tex]2^n=8[/tex]
[tex]2^n=2^3[/tex]
[tex]n=3[/tex]
Thus, the number of half-life periods will be, 3
The rate of change of speed of the belt is given by 0.06(10 - t) m/s^2, where t is in seconds. The speed of the belt is 0.8 m/s at t = 0. When the normal acceleration of a point in contact with the pulley is 40 m/s^2, determine (a) the speed of the belt; (b) the time required to reach that speed; and (c) the distance traveled by the belt.
Answer:
a) speed of belt = 0.8114m/s
b) time required = 0.02secs
c) distance traveled = 0.016m
Explanation:
The detailed and step by step calculation is as shown in the attached files.
A man (weighing 763 N) stands on a long railroad flatcar (weighing 3513 N) as it rolls at 19.8 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 4.68 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?
Answer:
0.8m/s
Explanation:
Weight of mas,F=763 N
Mass of man=[tex]\frac{F}{g}=\frac{763}{9.8}=77.86 kg[/tex]
By using [tex]g=9.8m/s^2[/tex]
Weight of flatcar=F'=3513 N
Mass of flatcar=[tex]\frac{3513}{9.8}=358.5 Kg[/tex]
Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg
Velocity of system=19.8m/s
Let v be the velocity of flatcar with respect to ground
Velocity of man relative to the flatcar=[tex]-4.68m/s[/tex]
Final velocity of man with respect to ground=v-4.68
By using law of conservation of momentum
Initial momentum=Momentum of car+momentum of flatcar
[tex]436.36(19.8)=77.86(v-4.68)+358.5v[/tex]
[tex]8639.928=77.86v-364.3848+358.5v[/tex]
[tex]8639.928+364.3848=436.36 v[/tex]
[tex]9004.3128=436.36v[/tex]
[tex]v=\frac{9004.3128}{436.36}[/tex]
[tex]v=20.6 m/s[/tex]
Initial speed of flatcar=Speed of system
Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s
A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseball travels a horizontal distance of 16.0 m and rotates through an angle of 44.5 rad. The baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?
Answer:
The tangential speed of a point on the "equator" of the baseball is 4.33 m/s.
Explanation:
Given that,
Linear speed of base ball = 42.5 m/s
Distance = 16.0 m
Angle = 44.5 rad
Radius of baseball = 3.67 cm
We need to calculate the flight time
Using formula of time
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{16.0}{42.5}[/tex]
[tex]t=0.376\ sec[/tex]
We need to calculate the number of rotation
Using formula of number of rotation
[tex]n=\theta\time 2\pi[/tex]
[tex]n=\dfrac{44.5}{2\pi}[/tex]
[tex]n=7.08[/tex]
We need to calculate the time for one rotation
Using formula of time
[tex]T=\dfrac{t}{n}[/tex]
Put the value into the formula
[tex]T=\dfrac{0.376}{7.08}[/tex]
[tex]T=0.053\ sec[/tex]
We need to calculate the circumference
Using formula of circumference
[tex]C=2\pi\times r[/tex]
Put the value into the formula
[tex]C=2\pi\times3.67\times10^{-2}[/tex]
[tex]C=0.23\ m[/tex]
The tangential speed is equal to the circumference divided by the time. it takes to complete one rotation.
We need to calculate the tangential speed
Using formula of tangential speed
[tex]v=\dfrac{C}{T}[/tex]
Put the value into the formula
[tex]v=\dfrac{0.23}{0.053}[/tex]
[tex]v=4.33\ m/s[/tex]
Hence, The tangential speed of a point on the "equator" of the baseball is 4.33 m/s.
Sound is more effectively transmitted into a stethoscope by direct contact than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope that has a sound gathering area of 15.0 cm2, and concentrates the sound onto two eardrums with a total area of 1.00 cm2 with an efficiency of 37.0%?
Answer:
Δβ = 28.2 dB
Explanation:
Attached is the explanation
A gas occupies 97 mL at 130 kPa. Find its volume at 225 kPa You must show all of your work to receive credit. .
Answer:
The answer to your question is 56 ml
Explanation:
Data
V1 = 97 ml
P1 = 130 kPa
V2 = ?
P2 = 225 kPa
Formula
Use Boyle's law to solve this problem because it relates to the volume and the pressure.
V1P1 = V2P2
Solve for V2
V2 = (V1P1) / P2
Substitution
V2 = (97 x 130) / 225
Simplification
V2 = 12610 / 225
Result
V2 = 56 ml
A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes. If the driver slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both these stops?
Answer:
For 24 seconds force exerted is 5092 N towards opposite direction of motion of bus.For 3.90 seconds force exerted is 31333 N towards opposite direction of motion of bus.Explanation:
We have equation of motion v = u + at
Initial velocity, u = 20 m/s
Final velocity, v = 0 m/s
Case 1:-
Time, t = 24 s
Substituting
v = u + at
0 = 20 + a x 24
a = -0.8333 m/s²
Force = Mass x Acceleration = 6110 x -0.8333 = -5092 N
Force exerted is 5092 N towards opposite direction of motion of bus.
Case 2:-
Time, t = 3.90 s
Substituting
v = u + at
0 = 20 + a x 3.90
a = -5.13 m/s²
Force = Mass x Acceleration = 6110 x -5.13 = -31333 N
Force exerted is 31333 N towards opposite direction of motion of bus.
The average force exerted on the bus with gentle braking is approximately -5092 N, and with slamming the brakes, it is approximately -31342 N. Negative values indicate the force direction is opposite to the motion.
Explanation:To calculate the average force exerted on the bus during both stops, we will use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by the acceleration (F = ma). Here, the acceleration can be calculated using the formula for deceleration since the bus is coming to a stop:
For gentle braking: Deceleration = (Final velocity - Initial velocity) / Time = (0 - 20.0 m/s) / 24.0 s = -0.8333 m/s2.For slamming the brakes: Deceleration = (Final velocity - Initial velocity) / Time = (0 - 20.0 m/s) / 3.90 s = -5.1282 m/s2.Now, we can calculate the average force (F) exerted in both cases by multiplying the mass of the bus (m = 6110 kg) by the deceleration (a).
Gentle braking average force: F = m × a = 6110 kg × -0.8333 m/s2 = -5092 N (approximately).Slamming the brakes average force: F = m × a = 6110 kg × -5.1282 m/s2 = -31342 N (approximately).Since the bus is stopping, the negative sign indicates the direction of the force is opposite to the initial direction of motion.
A stuntman with a mass of 82.5 kg swings across a pool of water from a rope that is 12.0 m. At the bottom of the swing the stuntman's speed is 8.65 m/s. The rope's breaking strength is 1,000 N. Will the stuntman make it across the pool without falling in?
Answer:
The stuntman will not make it
Explanation:
At the bottom of the swing, the equation of the forces acting on the stuntman is:
[tex]T-mg = m\frac{v^2}{r}[/tex]
where:
T is the tension in the rope (upward)
mg is the weight of the man (downward), where
m = 82.5 kg is his mass
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]m\frac{v^2}{r}[/tex] is the centripetal force, where
v = 8.65 m/s is the speed of the man
r = 12.0 m is the radius of the circule (the length of the rope)
Solving for T, we find the tension in the rope:
[tex]T=mg+m\frac{v^2}{r}=(82.5)(9.8)+(82.5)\frac{8.65^2}{12.0}=1322 N[/tex]
Since the rope's breaking strength is 1000 N, the stuntman will not make it.
What does the slope of gravitational potential energy vs velocity^2 graph represent?
Explanation:
Mechanical energy (the sum of potential and kinetic energy) is constant:
ME = PE + KE
ME = PE + ½ mv²
PE = ME − ½ mv²
So the slope of the line is -½ of the mass.
In a gravitational potential energy vs velocity^2 graph, the slope symbolizes the mass of the object. The graph represents the conversion of energies - kinetic and gravitational throughout a motion course. Peaks correspond to max potential energy and min kinetic energy, and vice-versa.
Explanation:The slope of the gravitational potential energy vs the square of the velocity graph represents the object's mass in specific physical circumstances. This is because the equation for kinetic energy (which this graph is showing an indirect relationship with) is (1/2)mv^2, where m is mass and v is velocity. The slope can be interpreted as mass due to the rearrangement of this formula into the form used for this graph (Potential energy = 1/2 * m * v^2), wherein the slope (m) symbolizes the mass of the object.
Moreover, the information encapsulated in the potential energy graph as a whole represents energy conversion from kinetic to gravitational and vice versa. At the peak of a slope/motion course, potential energy is at its maximum and kinetic energy at its minimum. The opposite occurs at the bottom of the slope: both kinetic energy and potential energy reach their minimum values.
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One of the many steps in the production of toothpaste is to screw the caps on the tubes, which is still a manual process, performed by on man, Mr. Bucket. Which statement about this situation is BEST?
A) This is most appropriate for an output measure of capacity.
B) In this case, the capacity of this step is not the maximum rate of output.
C) This is most appropriate for an input measure of capacity.
D) Utilization of the worker at this process step cannot be measures as it is a manual process.
Answer:A) This is most appropriate for an output measure of capacity.
Explanation: Toothpaste is a paste used to promote oral health and hygiene, Most toothpaste contain an active agent(floride) and it acts as an abrasive agent which helps to remove dental plaques or unwanted attachments to the teeth.
Among the options the Option(A) which says that THIS IS MOST APPROPRIATE FOR THE MEASURE OF CAPACITY is the statement that best Describes the situation. Toothpastes have a long history as it can be traced to about 4000years ago.
an electron aquires 3.45 e-16j of kinetic energy when it is accelerated by an electric field from plate a to b in a computer monitor what is the potential difference between the plates and which plate has the higher potential?
Answer:
Potential difference will be equal to 2156.25 volt
Explanation:
We have given kinetic energy of electron [tex]KE=3.45\times 10^{-16}J[/tex]
Charge on electron [tex]e=1.6\times 10^{-19}C[/tex]
Let the potential difference is V
So energy electron will be equal to [tex]E=qV[/tex], here q is charge on electron and V is potential difference
This energy will be equal to kinetic energy of the electron
So [tex]1.6\times 10^{-19}\times V=3.45\times 10^{-16}[/tex]
V = 2156.25 volt
So potential difference will be equal to 2156.25 volt
The decibel level of an orchestra is 90 db, and the violin section achieves a level of 80 dB. How does the sound intensity from the full orchestra compare to that from the violin section alone?
Final answer:
The sound intensity of the full orchestra is ten times greater than that of the violin section alone because a 10 dB increase corresponds to a tenfold increase in intensity.
Explanation:
The decibel level (dB) measures sound intensity on a logarithmic scale, where each 10 dB increase represents a tenfold increase in intensity. To compare the sound intensity of the full orchestra at 90 dB to the violin section at 80 dB, we can use the fact that an increase of 10 dB corresponds to a tenfold increase in intensity. Therefore, the full orchestra's sound intensity is ten times greater than that of the violin section alone.
This simplifies to 10, meaning that the sound intensity from the full orchestra is 10 times greater than that from the violin section alone.
In simpler terms, when comparing sound intensities, every 10 dB increase corresponds to a tenfold increase in intensity. Therefore, the difference of 10 dB between the full orchestra and the violin section results in the orchestra being 10 times louder in terms of sound intensity.
The sound intensity from the full orchestra is [tex]\( {10} \)[/tex] times greater than the sound intensity from the violin section alone.
To compare the sound intensities of the full orchestra and the violin section, we need to understand the relationship between decibels (dB) and sound intensity [tex]\( I \)[/tex].
The decibel scale is logarithmic and relates to sound intensity [tex]\( I \)[/tex] in the following way:
[tex]\[ \text{dB} = 10 \log_{10} \left( \frac{I}{I_0} \right) \][/tex]
where [tex]\( I \)[/tex] is the sound intensity of interest and [tex]\( I_0 \)[/tex] is the reference intensity (typically [tex]\( I_0 = 10^{-12} \)[/tex] W/m[tex]\(^2\))[/tex].
Step 1: Convert dB to intensity ratio
Given:
- Orchestra sound level [tex]\( L_{\text{orchestra}} = 90 \)[/tex] dB
- Violin section sound level [tex]\( L_{\text{violin}} = 80 \)[/tex] dB
The difference in decibel levels between the orchestra and the violin section gives us the intensity ratio:
[tex]\[ L_{\text{orchestra}} - L_{\text{violin}} = 90 \, \text{dB} - 80 \, \text{dB} = 10 \, \text{dB} \][/tex]
The intensity ratio in terms of decibels is related by:
[tex]\[ 10 \log_{10} \left( \frac{I_{\text{orchestra}}}{I_0} \right) - 10 \log_{10} \left( \frac{I_{\text{violin}}}{I_0} \right) = 10 \][/tex]
Step 2: Calculate the ratio of sound intensities
To find [tex]\( \frac{I_{\text{orchestra}}}{I_{\text{violin}}} \)[/tex]:
[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10^{10 / 10} \][/tex]
[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10^{1} \][/tex]
[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10 \][/tex]
Force is defined as mass times acceleration. Starting with SI base units, derive a unit for force. Using SI prefixes suggest a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour. (Assume 1 second deceleration time)
Force is derived from the equation F = ma, resulting in units of Newtons (N). For a 10-ton truck decelerating over 1 second from 55 mph, we convert mass to kg and speed to m/s to calculate force, likely resulting in a force measured in kilonewtons (kN) or meganewtons (MN) due to the large mass and deceleration involved.
Explanation:The subject of this question is how to derive a unit for force starting with SI base units, and finding a convenient unit for the force resulting from a specific collision scenario. Starting from the base, force (F) can be defined using Newton's second law, F = ma (force equals mass times acceleration), where mass has units of kilograms (kg), and acceleration is measured in meters per second squared (m/s²). Thus, the unit of force in the SI system is the Newton (N), defined as the force needed to accelerate a 1-kg object by 1 m/s².
To find a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour and assuming a 1 second deceleration time, we first convert the truck's mass to kilograms (10 tons = 9,071.85 kg, assuming 1 ton = 907.185 kg) and its speed to meters per second (55 mph ≈ 24.5872 m/s, assuming 1 mile = 1.60934 km and 1 hour = 3600 seconds). The deceleration rate (a) can be calculated assuming the final velocity (v) is zero and the time (τ) is 1 second, giving us a = -24.5872 m/s² (negative sign denotes deceleration). Thus, applying F = ma, the force of the collision can be calculated, and due to the large numbers involved, units such as kilonewtons (kN) or meganewtons (MN) may be more convenient depending on the exact outcome of the calculation.
The force resulting from the collision with a 10-ton trailer truck moving at 55 miles per hour is approximately 245.87 kN.
Starting with the equation for force, F = ma, where:
- F represents force,
- m represents mass,
- a represents acceleration.
In the International System of Units (SI), the base units are:
- Mass ( m ) is measured in kilograms (kg).
- Acceleration ( a ) is measured in meters per second squared [tex](\( \mathrm{m/s^2} \)).[/tex]
Therefore, the unit of force ( F ) in the SI system is:
[tex]\[ \mathrm{kg \cdot m/s^2} \][/tex]
This unit is commonly known as the Newton (N).
Now, let's calculate the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour with a deceleration time of 1 second.
First, we need to convert the mass of the trailer truck from tons to kilograms:
[tex]\[ 1 \text{ ton} = 1000 \text{ kg} \][/tex]
So, [tex]\( 10 \text{ tons} = 10,000 \text{ kg} \).[/tex]
Next, let's convert the speed from miles per hour to meters per second:
[tex]\[ 1 \text{ mile} = 1609.34 \text{ meters} \][/tex]
[tex]\[ 1 \text{ hour} = 3600 \text{ seconds} \][/tex]
So, [tex]\( 55 \text{ miles per hour} \)[/tex] is equivalent to:
[tex]\[ \frac{55 \times 1609.34}{3600} \text{ meters per second} \approx 24.587 \text{ m/s} \][/tex]
Given that deceleration time ( t ) is [tex]\( 1 \text{ second} \)[/tex], we can use the formula for acceleration:
[tex]\[ a = \frac{v}{t} \][/tex]
where v is the change in velocity.
Since the truck is decelerating, the change in velocity is from [tex]\( 24.587 \text{ m/s} \) to \( 0 \text{ m/s} \).[/tex]
[tex]\[ a = \frac{0 - 24.587}{1} \text{ m/s}^2 = -24.587 \text{ m/s}^2 \][/tex]
Now, we can calculate the force:
[tex]\[ F = m \times a = 10,000 \text{ kg} \times (-24.587 \text{ m/s}^2) \][/tex]
[tex]\[ F \approx -245,870 \text{ N} \][/tex]
Since force is a vector quantity and its direction is opposite to the direction of motion, the negative sign indicates that the force is acting in the opposite direction of the truck's motion.
For convenience, we can express this force in kilonewtons ( kN ):
[tex]\[ -245,870 \text{ N} = -245.87 \text{ kN} \][/tex]
So, the suggested convenient unit for the force resulting from the collision is [tex]$\mathrm{kN}$[/tex].
Complete Question:
Force is defined as mass times acceleration. Starting with SI base units, derive a unit for force. Using SI prefixes suggest a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour. (Assume 1 second deceleration time)
a. MN
b. GN
c. mN
d. kN
e. TN
Solar wind particles can be captured by the Earth's magnetosphere. When these particles spiral down along the magnetic field into the atmosphere, they are responsible for ?
Answer:
Solar wind particles can be captured by the Earth's magnetosphere. When these particles spiral down along the magnetic field into the atmosphere, they are responsible for?
Answer is Aurorae.
Explanation:
Solar Wind:
A planet's magnetic filed forms a shield to protect the surface of planet from energetic and charged particles coming from sun and other planets. The sun is continuously sending out charged particles , called Solar wind.
Magnetosphere of planet:
If a planet has a magnetic field then it will interact with the solar wind and deflect the charged particles in the wind. Due to which an elongated cavity in solar wind formed. This cavity is called magnetosphere of the planet.
When Solar wind particles run in a magnetic field , they are deflected and spiral down along magnetic field lines into the atmosphere.
Most of the solar wind particles deflected around the planet but a few particles manage to leak into the magnetic filed and become trapped in the magnetic field of the planet, to create Radiation Belts or Charged Particles Belts.
Variable Solar Wind can give some radiation belt particles with enough energy to spiral down along the magnetic filed into the atmosphere and Create Aurorae
Here Aurorae is an atmospheric phenomenon consisting of Bands, streamers of light, usually yellow , green or red , that move across the sky in the polar regions.
The charged particles from the solar wind spiral down along the Earth's magnetic field lines toward the poles, causing the atmospheric gases they collide with to emit light, resulting in auroras, commonly known as northern and southern lights.
Explanation:When charged particles, known as the solar wind, emitted from the Sun encounter the Earth's magnetosphere, they can get trapped and follow the magnetic field lines towards the Earth's poles. Upon interacting with atmospheric gases, these charged particles cause a natural light display in the sky known as the auroras. These spectacular light shows are called the aurora borealis, or northern lights, in the Northern Hemisphere, and the aurora australis, or southern lights, in the Southern Hemisphere.
These auroras are more than just beautiful; they represent the interaction between the Sun's energy and our planet's protective magnetic field. During periods of intense solar activity, the solar wind is stronger, leading to more significant and vibrant auroral displays. These geomagnetic storms caused by solar activity can also have consequential effects on power grids, satellite operations, and communication systems.
Two mountain bikers are rolling down a hill. They both have the same mass, but one is taller and has more air drag (friction, etc.) than the shorter one. Which will reach the bottom first?
Answer:The Taller one
Explanation: Air drag(friction) also known as air resistance,the opposing force acting against the force of gravity, if a falling Object have a high air drag it will mean the object have a high air resistance. The higher the air resistance the higher the impact of the force of gravity on an object. The force of gravity will act to want to overcome the opposing force in order to bring the Object or the taller person down.
In a situation with no energy losses, an object sliding without rolling will reach the bottom of an incline first since all its energy goes into descending speed. For two mountain bikers with the same mass, the one with less air resistance will reach the bottom first.
The question concerns the concept of rolling motion versus sliding motion and its effect on the speed at which objects descend an incline. In an ideal scenario with no energy losses, an object that is rolling will have some of its energy in rotational kinetic energy in contrast to an object that is sliding which will only have translational kinetic energy. As a result, the object sliding without rolling, given the same initial conditions, will reach the bottom first due to having all of its energy contributing to its descent speed.
In the case of two mountain bikers with the same mass but differing in air resistance, the biker with less air drag will reach the bottom first. Air resistance acts as a form of friction that takes away some of the energy that could otherwise be converted into speed, thereby slowing down the biker.
A 16 V battery does 1705 J of work transferring charge. How much charge is transferred? Answer in units of C.
Answer:
Explanation:
Given
Voltage [tex]V=16\ V[/tex]
Work done [tex]W=1705\ J[/tex]
Work is Equivalent to energy
We know that Charge is given by
[tex]Q=I\cdot t[/tex]
where I=current
t=time
Energy [tex]E=P\times t[/tex]
P=power
P=V\times I
V=Voltage
I=current
Also Energy [tex]E=V\cdot I\cdot t[/tex]
[tex]I=\frac{E}{V\cdot t}[/tex]
Substitute the value of I in charge
[tex]Q=\frac{E}{V\cdot t}\times t[/tex]
[tex]Q=\frac{E}{V}[/tex]
[tex]Q=\frac{1705}{16}[/tex]
[tex]Q=106.56\ C[/tex]
In a nuclear power plant, the temperature of the water in the reactor is above 100°C because of what?
Answer:
The temperature of the water increases because the nuclear reactor heats it producing steam
Explanation:
The nuclear power plants are usually defined as those thermal plants where the nuclear reactors are used in order to generate heat that eventually leads to the rotating of the turbines and produces electricity. Here the nuclear reactor heats the water, and it increases above a temperature of 100°C, where this heat energy plays a key role in the entire process. It is an efficient method as it does not lead to the emission of any green house gases that are harmful to the environment.
The liquid in one container drops 100 F, while the same liquid in a different container drops 100C. How does the change in thermal energy in the two compare? Explain your answer.
Answer:
Explanation:
The thermal energy is given as
E=mc(T2-T1)
For the temperature change.
ΔT can be expressed in units of Kelvin or degrees Celsius. The ΔT of Kelvin is equal to that of ΔT of Celsius but not equal to the ΔT of Fahrenheit.
Therefore change in temperature of 100F is not equal to change in temperature of 100C
°F= 9/5 °C +32
So let assume 20°C, so the increase of 100°C will give 120°C.
Then °F = 68°F
Now the equivalent of 20°C is 68°F.
So let see the value of 120°C, which is the increase given.
°F= 9/5 ×120+32
°F= 248°F
The change in temp in Fahrenheit is 248-68=180°F
1°F change is = to 0.56°C change
Therefore, 100°F change in temperature in Fahrenheit is equal to 56°C in Celsius
Therefore the thermal energy is not the same, since they have are the same mass and specific heat but different changer in temperature.
So using the formulae given above, the thermal energy of the 100F change is greater than that of the 100C change..
A temperature drop of 100°C corresponds to a larger change in thermal energy than a drop of 100°F because 100°C equates to a more substantial temperature change (around 180°F). The container with the 100°C temperature drop, therefore, experiences a greater thermal energy change.
Explanation:In comparing the change in thermal energy of liquids with a temperature drop of 100°F and 100°C, understanding that 1°C is equal to 1.8°F is crucial. To compare the energy change between the two, we must first recognize that the change of 100°F is approximately equivalent to a change of 55.5°C (100°F / 1.8 = 55.5°C).
Since the specific heat capacity is a property of a material that indicates how much heat energy it requires to change its temperature by a single degree, and this value is typically given in J/g°C, it is clear that a larger temperature change in Celsius would require more energy. Therefore, the container experiencing a drop of 100°C will require the removal of more thermal energy compared to the container with a drop of 100°F.